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  "body": "\n## Statement\n\nLet $0 \u003c \\delta \u003c 1$, and let $C \u003e 0$, $P \\in \\R$ and $0 \\le A \u003c \\pi$.\nLet $\\varphi$ be a [[holomorphic|0NAHXR1FRH]] function on an open set containing the closed half-plane $H = \\set{v + iw : v \\ge -\\delta}$, such that\n$$\\abs{\\varphi(v + iw)} \\le C e^{P v + A \\abs{w}} \\qquad \\text{for all } v + iw \\in H.$$\nThen the series $f(x) = \\sum_{k=0}^{\\infty} \\varphi(k) (-x)^k$ converges for $0 \u003c x \u003c e^{-P}$, and $f$ has an [[analytic continuation|0M3YQJ81NB]] along $(0, \\infty)$.\nFor every $s \\in \\C$ with $0 \u003c \\operatorname{Re} s \u003c \\delta$, the [[Mellin transform|N5PVGRJE1T]] of this continuation converges and\n$$\\int_0^\\infty x^{s-1} f(x) \\, dx = \\frac{\\pi}{\\sin(\\pi s)} \\, \\varphi(-s).$$\n\n## Remarks\n\nThis is the form of the theorem proved by G. H. Hardy.\nThe idea is to write $f$ as a Mellin–Barnes integral: by the residue theorem, $f(x) = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} \\frac{\\pi}{\\sin(\\pi s)} \\varphi(-s) x^{-s} \\, ds$ for $0 \u003c c \u003c \\delta$, since $\\pi / \\sin(\\pi s)$ has residue $(-1)^k$ at $s = -k$.\nThe growth bound with $A \u003c \\pi$ beats the decay $e^{-\\pi \\abs{w}}$ of $1/\\sin(\\pi s)$ and makes the integral converge; Mellin inversion then gives the formula.\nWriting $\\varphi(k) = \\lambda(k) / \\Gamma(k + 1)$ and using $\\Gamma(s)\\Gamma(1 - s) = \\pi / \\sin(\\pi s)$ gives the classical form: if $f(x) = \\sum_{k \\ge 0} \\lambda(k) (-x)^k / k!$, then $\\int_0^\\infty x^{s-1} f(x) \\, dx = \\Gamma(s) \\lambda(-s)$, under the corresponding conditions on $\\lambda$.\nExample: $\\varphi = 1$ gives $f(x) = 1/(1 + x)$ and $\\int_0^\\infty x^{s-1}/(1 + x) \\, dx = \\pi / \\sin(\\pi s)$ for $0 \u003c \\operatorname{Re} s \u003c 1$.\nExample: $\\lambda = 1$ gives $f(x) = e^{-x}$ and the formula $\\int_0^\\infty x^{s-1} e^{-x} \\, dx = \\Gamma(s)$.\nThe growth condition is needed: $\\varphi(z) = \\sin(\\pi z)$ vanishes at all integers, so $f = 0$, but $\\varphi(-s) \\ne 0$ for non-integer $s$; here $A = \\pi$ and the bound with $A \u003c \\pi$ fails.\n\n## Explanations\n\n### L1 human\n\nThe theorem computes the Mellin transform of a function from the coefficients of its power series.\nIf the $k$-th coefficient is $\\varphi(k)$ times $(-1)^k$, then the transform is essentially $\\varphi$ evaluated at $-s$.\nSo a formula valid for the whole numbers $k = 0, 1, 2, \\dots$ is pushed to negative and complex values.\nRamanujan used it, without a full proof, to evaluate a great many definite integrals.\n\n### L2 human\n\nThe growth conditions make the passage from integers to all complex numbers legitimate.\nBy Carlson's theorem, a function of this type is determined by its values at the non-negative integers.\nThe proof is an instance of the general Mellin method.\nA Mellin–Barnes integral, shifted to the left, gives the power series at $0$, and its inverse transform is the right side.\nThe theorem yields at once the Mellin transforms of the binomial series $(1 + x)^{-a}$, of Bessel functions and of many hypergeometric functions.\n",
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